How do we show that $(2k+1)^{2n} \bmod 4 = 1$?

Question

In a recent (March 24, 2018) [video][1], Mathologer tells us that "for an even power of an odd number [all divided by $4$], the only possible remainder is $1$." He leaves the proof as an "easy puzzle."

With the help of my cat, I've rendered his claim thus: $$(2k+1)^{2n} \bmod 4 = 1$$ where $k$ and $n$ are non-negative integers.

I can show that this is true for $n$ = 1 ($n$ = 0 as well, but that's trivial):

$$(2k+1)^2 = 4k^2 + 4k +1= 4(k^2+k) +1$$ Since $k$ is an integer, $k^2+k$ is also an integer, which I can express as $m$. So:

$$(2k+1)^2 = 4m +1$$ and $$(4m+1) \bmod 4 = 1$$

How do I go about generalizing this for $n > 1$? Induction? Binomial theorem? Am I pointed in the right direction?

Answer 1

You have

$$(2k + 1)^{2n} = \sum_{j=0}^{2n} {2n\choose k} 4^jk^j = 1 + \sum_{k-1}^{2n}{2n\choose k}4^jk^j.$$

'Nuff said?

Answer 2

$$(2k+1)^2=4k^2+4k+1\equiv 1\pmod 4$$

$$(2k+1)^{2n}\equiv 1^n \equiv 1 \pmod 4$$

Answer 3

Use the fact that $$a = b\mod r$$ implies $$ka = kb \mod r$$ And make an induction on $n$.

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Previous answer:

Note that the residue of a odd number is either $1$ or $3$ in $\mathbb{Z}/4\mathbb{Z}$, and $1^2 = 3^2 = 1$.

Answer 4

HINT: Use $a \bmod m = b \Rightarrow a^n \bmod m =b^n$

You have already proved that $(2k+1 )^2 \bmod 4 =1$, so you can easily conclude it.

And in case you need the proof:

Prove that $a \bmod m = b \Rightarrow a^n \bmod m =b^n$

We have $a-b$ is divisible by $m$.

Then $(a-b)(a^{n-1}b^{0}+a^{n-2}b^{1}+a^{n-3}b^2+...+a^{1}b^{n-2}+a^{0}b^{n-1})$ must also be divisible by $m$.

Symplify the expression above, we will get $a^n-b^n$ is divisible by $m$.